Let
$$
u=1+\frac{x^3}{3!}+\frac{x^6}{6!}+\cdots
$$
$$
v=x+\frac{x^4}{4!}+\frac{x^7}{7!}+\cdots
$$
$$
w=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+\cdots
$$
Show that $$u^3+v^3+w^3-3uvw=1$$ It turns out to be an interesting application of the 3rd root of unity, which greatly simplified the (could be) tedious calculation. I wonder if it has any deeper interpretation. (At least I don't see how to easily generalize it.) Can anybody explain this? (I don't need help on solving this problem.)

There are more relations.If $u(x)=\sum \limits_{k=0}^\infty \frac{x^{3k}}{(3k)!}$ then $u(2x)=3u^2(x)-2u(-x)$ and also $u(3x)=1+9u(x)v(x)w(x)$ .For more details, you can see my question math.stackexchange.com/questions/180832/…
–
MathloverNov 29 '12 at 6:37

So $u^3 + v^3 + w^3 - 3uvw$ is constant with respect to $x$ (since its derivative is always $0$), and evaluating it at any particular value of $x$ gives its value for all $x$. Now choosing $x=0$, we get $u = 1$, $v=w=0$, and so $u^3 + v^3 + w^3 - 3uvw = 1$.